channel.invokeMethod() 在 FirebaseMessaging.onBackgroundMessage() 中不起作用；

Question

在颤振中 在 a 中调用像这样的原生方法

_channel.invokeMethod()

FirebaseMessaging.onBackgroundMessage(_firebaseMessagingBackgroundHandler);

例外

[ERROR:flutter/runtime/dart_vm_initializer.cc(41)] Unhandled Exception: MissingPluginException(No implementation found for method wakeupscreen on channel flutter_native)

这是我的 _firebaseMessagingBackgroundHandler()

@pragma('vm:entry-point')
Future<void> _firebaseMessagingBackgroundHandler(RemoteMessage message) async {
  print("Handling a background message: ${message.messageId}");
  await Firebase
      .initializeApp(); //make sure firebase is initialized before using it (showCallkitIncoming)

  wakeupscreen();

}

// wakeupscreen() func -- this is a global function so does not require any class instance to run

wakeupscreen() async {
  try {
    const platform = MethodChannel('flutter_native');
    await platform.invokeMethod('wakeupscreen');
  } on PlatformException catch (e) {
    print("Failed to invoke the method: '${e.message}'.");
  }
}

我尝试在普通代码中使用相同的wakeupscreen()函数：

InkWell(
                  onTap: () {
                    Future.delayed(Duration(seconds: 5), () {
                      wakeupscreen();
                    });
                  },
                  child: Text("wakeup in 5 sec"),
                ),

正常调用时，它可以完美地工作，但是一旦在

_firebaseMessagingBackgroundHandler

 中调用相同的函数，它就会抛出异常。

Answer 1

在main.dart中，设置一个方法通道来处理来自后台isolate的消息：

// Inside main.dart

MethodChannel _channel = MethodChannel('flutter_native');

// Set up method channel to communicate with background isolate
_channel.setMethodCallHandler((call) async {
  if (call.method == 'wakeupscreen') {
    wakeupscreen(); // Call your wakeupscreen function here
  }
});

从wakeupscreen函数中删除方法通道初始化：

// Remove this line from the wakeupscreen function
const platform = MethodChannel('flutter_native');

这些更改确保方法通道被正确配置为管理来自后台隔离的消息，并确保根据需要调用wakeupscreen函数。